TITLE: Pi, the Lemniscatic elliptic functions, and the Dixonian elliptic functions
QUESTION [8 upvotes]: $n=2$  

Circle: $x^2+y^2=c^2$. We all know the role of $\pi$ in the circle and the trigonometric functions,
$$\pi_2 = \color{brown}{B\big(\tfrac12,\tfrac12\big)}=3.1415\dots$$
with the beta function $B(x,y)$ and circumference for $D=1$
$$C=\pi_2 = 3.1415\dots$$

$n=4$  

Lemniscate: $(x^2+y^2)^2=2c^2(x^2-y^2)$ (or $r^2=2c^2\cos2\theta$ in polar form):
$\hskip1.4in$
The lemniscate constant $L$ plays an analogous role to $\pi$,
$$L =\frac{\color{brown}{B\big(\tfrac14,\tfrac14\big)}}{2\sqrt2}= \frac{B\big(\tfrac12,\tfrac14\big)}{2}=  \frac{\sqrt\pi\,\Gamma\big(\tfrac14\big)}{2\Gamma\big(\tfrac34\big)}=\frac{\Gamma^2\big(\tfrac14\big)}{2\sqrt{2\pi}}=2.622057\dots$$
and if $c=1$,
$$\text{Arclength} = 2L = 5.2441\dots$$
The minimal half period $\omega_1$ of the lemniscatic elliptic functions is,
$$\omega_1=\frac{L}{\sqrt2}=\frac{\Gamma^2\big(\tfrac14\big)}{4\sqrt{\pi}}$$

$n=3$

Object?: ($3$rd or $6$th deg poly?)  However, it seems we missed a step,
$$\pi_3 = \color{brown}{B\big(\tfrac13,\tfrac13\big)}=2^{1/3}\,B\big(\tfrac12,\tfrac13\big) =\frac{\Gamma^2\big(\tfrac13\big)}{\Gamma\big(\tfrac23\big)}=\frac{\sqrt{3}\,\Gamma^3\big(\tfrac13\big)}{2\pi}=5.29991\dots$$
which conveniently is the fundamental constant of the Dixonian elliptic functions.

Q: So does the constant $\pi_3$ play an analogous role for some geometric object akin to the circle and lemniscate?

Edit: To clarify (courtesy of reuns' comments): to know the arc length of a circle you need $\pi_2$. For a lemniscate, you need  $L$. So is there a simple geometric object where to know its arc length you need  $\pi_3$?

REPLY [4 votes]: Langer and Singer, in their paper, present the trefoil, which has the Cartesian equation
$$(x^2+y^2)^3=2(x^3-3xy^2)$$ 
and polar equation $r^3=2\cos3\theta$:
$\hskip1.3in$
Using the normal formula for determining the arclength of a polar curve, its total arclength comes out to
$$\text{Arclength} =\sqrt{3}\,\pi_3=9.179724\dots$$
Additionally, they also present a unit-speed parametrization in terms of the Dixon functions. Letting $\sigma=\operatorname{sm}\dfrac{s}{\sqrt3}$ and $\chi=\operatorname{cm}\dfrac{s}{\sqrt3}$, we have the arclength parametrization
$$f(s)=\begin{pmatrix}\dfrac{3\chi\sigma^4}{2(\chi^3\sigma^3-1)}\\\dfrac{\sqrt{3}\chi\sigma(1+\chi^3)}{2(1-\chi^3\sigma^3)}\end{pmatrix}$$
See this related paper as well.